Conjugation action on the classifying space of circle Let $T$ denote the unit sphere in complex plane. $\mathbb{Z}/2$ acts by complex conjugation on $T$. There is an induced action on $BT$. The cohomology of the homotopy orbit space of this action is
$$
H^*(BT//(\mathbb{Z}/2),\mathbb{F}_2) = H^*(B\mathbb{Z}/2,\mathbb{F}_2)\otimes H^*(BT,\mathbb{F}_2)=\mathbb{F}_2[t,x]
$$
where degrees are given by $|t|=1$ and $|x|=2$. This is because the associated fibration has a section. But I am curious about the structure of the cohomology ring as a module over the Steenrod algebra. 
 A: It is the interesting one, with $Sq^1(x) = tx$ and the remainder determined by the axioms of Steenrod operations. This may be seen as the homotopy orbits is a model for BO(2), where one knows the Steenrod operations. (Here I am assuming the choice of class $x$ which pulls back to zero under the section; under the equivalence to $BO(2)$ it then bevomes the second Stiefel--Whitney class.)
Alternatively, you can find the mapping torus $T$ of the complex conjugation map on $\mathbb{CP}^1$ as a natural subspace of $BT /\!\!/ \mathbb{Z}/2$, and the fundamental class of $T$ pairs to 1 with $tx$. By naturality, the question now becomes whether
$$\mathrm{Sq}^1 : H^2(T ; \mathbb{F}_2) = \mathbb{F}_2\{x\} \to H^3(T ; \mathbb{F}_2) = \mathbb{F}_2\{tx\}$$
is non-zero or not. But $T$ is a manifold, and Steenrod-squaring into the top degree is always given by cup product with the corresponding Wu class, in this case $v_1$. As the total Stiefel--Whitney class and the total Wu class are related by $w = \mathrm{Sq}(v)$, we have $v_1 = w_1$, so the question becomes to identify the first Stiefel--Whitney class of the 3-manifold $T$. But $T$ is clearly nonorientable, as it is the mapping torus of an orientation-reversing diffeomorphism, so $w_1 \neq 0$; thus $w_1$ must be $t$, so $\mathrm{Sq}^1(x) = t \cdot x$.
